Thevenin's theorem

 Thevenin's theorem :

This method is used for the reduction of complex circuits into simpler ones.

According to this theorem, any two-terminal network, containing two or more sources can be replaced by an equivalent network that contains equivalent Thevenin voltage Vth and Thevenin resistor Rth in series. 

This network can be replaced like this equivalent circuit shown below.

Any linear circuit having more number of active and passive elements can be replaced by the equivalent circuit which contains Thevenin's voltage and resistor.

Thevenin theorem is applicable for linear and bilateral networks.

This theorem does not apply to diodes and transistors.

To solve the circuits using Thevenin's theorem we need to follow two steps to find Vth and Rth.

Step 1: In any linear circuit, if we need to find the equivalent voltage we need to remove the load resistor to find the voltage across the given terminal.

Step 2: To find the equivalent resistance we need to short circuit the independent voltage source and open-circuited the independent current source if we have any dependent source then we leave it simply. Then we will calculate the Rth.

Example Problems:

Determine the Thevenin equivalent voltage and resistance in the following circuit across terminals A and B. Also, determine the load current and the load voltage.

To find Vth: Firstly we need to remove the resistor across the terminal AB.

Therefore the circuit will be like as in the figure shown below.

Now we will calculate the Thevenin voltage.

Therefore applying Kirchoff's voltage law we get,

By calculating the open circuit voltage across the terminal AB we get,

                                            Vth = I * 3kΩ

                                                        = 1.25mA * 3000

                                                 Vth = 3.75 V

To find Rth we need to short circuit the voltage source,

So we get,

Now calculate the Rth.

Here 5kΩ and 3kΩ are connected in parallel.

Now 15k/8 Ω and 2kΩ are connected in series. So we get,

Therefore the new circuit looks like